দেওয়া আছে, $$ \begin{aligned} \vec{A}&=(x^2yz)\hat{i}-(2xz^3)\hat{j}+(xz^2)\hat{k} \\ \vec{B}&=(2z)\hat{i}+y\hat{j}-x^2\hat{k} \end{aligned} $$ প্রথমে $\vec{A} \times \vec{B}$ নির্ণয় করি: $$ \begin{aligned} \vec{A} \times \vec{B}&=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x^2yz & -2xz^3 & xz^2 \\ 2z & y & -x^2 \end{vmatrix} \\ &=\hat{i}((-2xz^3)(-x^2) - (xz^2)(y)) - \hat{j}((x^2yz)(-x^2) - (xz^2)(2z)) + \hat{k}((x^2yz)(y) - (-2xz^3)(2z)) \\ &=\hat{i}(2x^3z^3 - xyz^2) - \hat{j}(-x^4yz - 2xz^4) + \hat{k}(x^2y^2z + 4xz^4) \\ &=(2x^3z^3 - xyz^2)\hat{i} + (x^4yz + 2xz^4)\hat{j} + (x^2y^2z + 4xz^4)\hat{k} \end{aligned} $$ এখন, $\frac{\partial}{\partial y}(\vec{A} \times \vec{B})$ নির্ণয় করি: $$ \begin{aligned} \frac{\partial}{\partial y}(\vec{A} \times \vec{B})&=\frac{\partial}{\partial y}(2x^3z^3 - xyz^2)\hat{i} + \frac{\partial}{\partial y}(x^4yz + 2xz^4)\hat{j} + \frac{\partial}{\partial y}(x^2y^2z + 4xz^4)\hat{k} \\ &=(-xz^2)\hat{i} + (x^4z)\hat{j} + (2x^2yz)\hat{k} \end{aligned} $$ এরপর, $\frac{\partial^2}{\partial x \partial y}(\vec{A} \times \vec{B})$ নির্ণয় করি: $$ \begin{aligned} \frac{\partial}{\partial x}\left[\frac{\partial}{\partial y}(\vec{A} \times \vec{B})\right]&=\frac{\partial}{\partial x}(-xz^2)\hat{i} + \frac{\partial}{\partial x}(x^4z)\hat{j} + \frac{\partial}{\partial x}(2x^2yz)\hat{k} \\ &=(-z^2)\hat{i} + (4x^3z)\hat{j} + (4xyz)\hat{k} \end{aligned} $$ এখন, $(1, 0, -2)$ বিন্দুতে মান বসিয়ে পাই: $$ \begin{aligned} \left. \frac{\partial^2}{\partial x \partial y}(\vec{A} \times \vec{B}) \right|_{(1, 0, -2)}&=-(-2)^2\hat{i} + 4(1)^3(-2)\hat{j} + 4(1)(0)(-2)\hat{k} \\ &=-4\hat{i} - 8\hat{j} + 0\hat{k} \\ &=-4\hat{i} - 8\hat{j} \end{aligned} $$ নির্ণেয় মান: $-4\hat{i} - 8\hat{j}$।